Lemma 15.12.2D0B5A6CE28E9

Proof. Since $u\sim \varphi $, the smaller-background envelope $P_{\varepsilon \theta }(u)$ is not identically $-\infty $. If $F=0$, then theorem 15.2.2, applied to $u$ and $u+1$ with coefficient zero, would give $u\geq u+1$; hence $N=\| F\| _ q>0$.

Apply lemma 15.3.1 in the semipositive big background $(1-\varepsilon )\theta $ to $F/N$, whose $L^ q$ norm is one. It gives a bounded $s$ with $-1\leq s\leq 0$ and a constant $b_ q>0$, depending only on this fixed background and on $q$, such that

Put $h=P_{\varepsilon \theta }(\varphi )+s$. This is a $\theta $-plurisubharmonic function with $h\preceq u$. Since $\theta _ h$ is the sum of the positive currents $\varepsilon \theta +\mathrm{dd}^{\mathrm{c}}P_{\varepsilon \theta }(\varphi )$ and $(1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}s$, the same density bound holds for $\mathrm{MA}^{\mathrm{ext}}_\theta (h)$. To justify this, we truncate $P_{\varepsilon \theta }(\varphi )$ at finite levels, use positivity of bounded mixed products, and exhaust the set where $P_{\varepsilon \theta }(\varphi )$ is finite; its pole set has zero volume.

If $N<b_ q$, we would have

Since $N/b_ q<1$, theorem 15.2.2 would then imply $u\geq h+t$ for every $t$, which is impossible at a point where both potentials are finite. Thus $N\geq b_ q$. The construction of $s$ used no property of $\varphi $, which proves the last assertion. $\square$